6.7 Uniqueness proofs

Technique 6.7.1

The basic method is to assume that there are two objects, e.g. aa and bb which are distinct (i.e. aba\neq b) and show that from this we can derive that a=ba=b (i.e. a contradiction).

Example 6.7.1

Show that if rr is an irrational number, there is a unique integer nn such that the distance between rr and nn is less than 12\frac{1}{2}.

There are two parts to this proof; the first is to prove that \saythere exists some object with the desired properties and the second is to show that this object is unique (it is of course possible to prove these in the opposite order, but I personally prefer to first prove existence as this can shine light on how to carry out the uniqueness proof).


It helps to first consider the problem informally. Usually with problems about numbers, it’s useful to draw a number line, such as this one.

We must plot rr somewhere. We know that rr is irrational, so anywhere between two integers will do (this is a sketch). For example, we might have


It would be nice to show that the distance between rr and one of the integer neighbours is less than 12\frac{1}{2}. We first need to a little intuition to mathematics translation. In our diagram we can label aa where aa is the greatest integer which is less than or equal to rr, i.e.


We can then construct the inequality

ar<a+1\displaystyle a\leqq r<a+1 (6.31)

As rr is irrational we cannot have r=ar=a as aa is rational, so

a<r<a+1\displaystyle a<r<a+1 (6.32)

Then, all that remains to show is that either

|ar|<12 or |(a+1)r|<12|a-r|<\frac{1}{2}\text{ or }|(a+1)-r|<\frac{1}{2} (6.33)

The only way that both of these can be false is if r=a+(a+1)2=2a+12r=\frac{a+(a+1)}{2}=\frac{2a+1}{2}, but this is rational, so rr cannot be equal to this. Therefore, such an nn (either n=an=a or n=a+1n=a+1) does exist.


This proof is by contradiction. First let us assume that there are two integer xyx\neq y which exist, and are different (i.e. xyx\neq y), and that

|xr|<12 and |yr|<12\left\lvert x-r\right\rvert<\frac{1}{2}\text{ and }\left\lvert y-r\right\rvert% <\frac{1}{2} (6.34)

We would like to show that x=yx=y. A standard technique to apply in such cases is to note that this is equivalent to xy=0x-y=0, and as we are interested in distances, we can show that

|xy|\displaystyle|x-y| =|xa+ay|\displaystyle=|x-a+a-y| (6.35)
|xa|+|ay|\displaystyle\leqq|x-a|+|a-y| By the triangle inequality (6.36)
<12+12\displaystyle<\frac{1}{2}+\frac{1}{2} By assumption in Equation 6.34 (6.37)
<1\displaystyle<1 (6.38)

As x,yx,y are integers, it must therefore be that x=yx=y, which contradicts our earlier hypothesis that xyx\neq y, so any such nn is unique.